Explain the expectation of white balls when randomly pick 5 balls in a box with 3 white balls and 6 red balls. Let $X$ denote the number of white balls, and $X=0,1,2,3$, so we can calculate $E(X)$:
By definition, we can find that $E(X)= \frac53$
But it also equal to $5 \cdot \frac39$.
I wonder how to explain the $5 \cdot \frac39$, because it looks like that 3 red balls dissolve in 9 balls uniformly, and every time we choose "one" ball, we will get $\frac39$ white ball, after 5 times we will get $5 \cdot \frac39$ white balls.
Is it an accidient or it make sense?
 A: Yes, it make sense. We can find expectation, using indicator method. Let
$$A_j = \left\{ \text{at the }j-\text{th step a white ball is drawn}  \right\}, \quad I_j=I_{A_j} =\begin{cases} 1, \text{if event }A_j \text{ happened}, \\ 0, \text{ otherwise}  .\end{cases}$$
Then
$$\mathbb{E} X = \mathbb{E} \sum\limits_{j=1}^{5} I_j = \sum\limits_{j=1}^{5} \mathbb{E}I_j = \sum\limits_{j=1}^{5} \mathbb{P}(A_j) = 5 \mathbb{P}(A_1)= 5\cdot \frac{3}{9},$$
where the second equality is true during linearity of expectation and last two equalities is true during symmetry.
A good question here is why $\mathbb{P}(A_j) = \frac{3}{9} \quad \forall j \in \left\{1,...,5 \right\}$. It follows from symmetry or we can explain it as @N. F. Taussig wrote or we can verify this, using conditional probability (but this take some time). Anyway, the probability $\mathbb{P}(A_2)$ can be calculated quickly
$$\mathbb{P}(A_2) = \mathbb{P}(A_2|A_1) \cdot \mathbb{P}(A_1) + \mathbb{P}(A_2|\overline{A_1})\cdot \mathbb{P}(\overline{A_1}) =$$
$$ = \frac{2}{8}\cdot \frac{3}{9} + \frac{3}{8}\cdot \frac{6}{9}   = \frac{6+18}{72}= \frac{1}{3}.$$
In general case
$$\mathbb{P}(A_{k+1})=\sum\limits_{s=0}^{k} \mathbb{P}(A_{k+1}|H_{ks})\cdot \mathbb{P}(H_{ks}),$$
where
$$H_{ks}=\left\{  s \text{ of the } k \text{ balls are white} \right\}.$$
So, we can easily find that
$$\mathbb{P}(H_{ks})=\frac{C_3^s\cdot C_6^{k-s}}{C_9^k}, \quad s=0,...,k, \quad \mathbb{P}(H_{ks})=0, \text{ if } s>3,$$
$$\mathbb{P}(A_{k+1}|H_{ks})=\frac{3-s}{9-k}.$$
Finally, we have
$$\mathbb{P}(A_{k+1})=\sum\limits_{s=0}^{k} \frac{C_3^s\cdot C_6^{k-s}}{C_9^k} \cdot 
\frac{3-s}{9-k}=\frac{3}{9} \cdot \sum\limits_{s=0}^{k} \frac{C_2^s\cdot C_6^{k-s}}{C_8^k}=\frac{3}{9}, \quad k=0,...,4. $$
